poj3020(最小边覆盖)

思路：要安放一下类似雷达之类的东西，相邻的一个点会被覆盖，但是，只能覆盖一个，不管是东西南北，那么另外一个被覆盖的点就不用在安放雷达了，求最少的安放雷达的数目；显然就是最小边覆盖问题；

对每个坐标表定一个固定的数字，然后根据题意建图；与这题一样的还有lightoj1152。

题目链接

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
#define MEM(a,b) memset(a,b,sizeof a)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
char gg[50][50];
int link[50 * 50];
vector<int> G[50 * 50];
int dx[] = {-1,0,1,0};
int dy[] = {0,-1,0,1};
bool vis[50 * 50];
int n,m;
bool dfs(int u){
	for (int i = 0;i < G[u].size();i++){
		int v = G[u][i];
		if (!vis[v]){
			vis[v] = true;
			if (link[v] == -1 || dfs(link[v])){
				link[v] = u;
				return true;
			}
		}
	}
	return false;
}

int Hungary(){
	MEM(link, -1);
	int ret = 0;
	for (int i = 1;i <= n * m;i++){
		MEM(vis, false);
		if (dfs(i)) ret++;
	}
	return ret;
}
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int T;
	cin >> T;
	while(T--){
		cin >> n >> m;
		for (int i = 1;i <= n * m;i++)
			G[i].clear();
		for (int i = 1;i <= n;i++){
			scanf("%s",gg[i] + 1);
		}
		int cnt = 0;
		for (int i = 1;i <= n;i++){
			for (int j = 1;j <= m;j++){
				if (gg[i][j] == '*'){
					cnt++;
					for (int k = 0;k < 4;k++){
						int nx = i + dx[k];
						int ny = j + dy[k];
						if (nx < 1 || nx > n || ny < 1 || ny > m) continue;
						if (gg[nx][ny] == '*')
							G[(i - 1) * m + j].push_back((nx - 1) * m + ny);
					}
				}
			}
		}
		printf("%d\n",cnt - Hungary() / 2);
	}
	return 0;
}

// #pragma comment(linker, "/STACK:1024000000,1024000000")
//Light_OJ 1152
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
// #define DEBUG
#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__ )
#else
#define debug(...)
#endif
#define MEM(x,y) memset(x, y,sizeof x)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n, m;
const int maxn = 30;
char s[maxn][maxn];
int link[maxn*maxn];
bool vis[maxn*maxn];
vector<int> G[maxn*maxn];
bool Search(int u){
	for (int i = 0;i < G[u].size();++i){
		int v = G[u][i];
		if (vis[v]) continue;
		vis[v] = true;
		if (link[v] == -1 || Search(link[v])){
			link[v] = u;
			return true;
		}
	}
	return false;
}
int Hungary(){
	int ans = 0;
	memset(link, -1,sizeof link);
	for (int i = 1;i <= n * m;++i){
		memset(vis, false,sizeof vis);
		if (Search(i)) ans++;
	}
	return ans;
}
int dx[] = {-1,0,1,0};
int dy[] = {0,-1,0,1};
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int t, icase = 0;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&m);
		for (int i = 1;i <= n;++i)
			scanf("%s",s[i] + 1);
		for (int i = 0;i <= n * m;++i)
			G[i].clear();
		int cnt = 0;
		for (int i = 1;i <= n;++i){
			for (int j = 1;j <= m;++j){
				if (s[i][j] == '*'){
					cnt++;
					for (int k = 0;k < 4;++k){
						int nx = i + dx[k];
						int ny = j + dy[k];
						if (nx < 1 || nx > n || ny < 1 || ny > m) continue;
						if (s[nx][ny] == '*'){
							G[(i-1)*m+j].push_back((nx-1)*m+ny);
						}
					}
				}
			}
		}
		printf("Case %d: %d\n", ++icase, cnt - Hungary()/2);
	}
	return 0;
}